Q. Solve the system of equations.y=10x−80y=x2−12x+41Write the coordinates in exact form. Simplify all fractions and radicals.(______,______)
Substitute y Equation: Substitute y from the first equation into the second equation. Since y=10x−80, we can replace y in the second equation y=x2−12x+41 with 10x−80. This gives us the equation 10x−80=x2−12x+41.
Rearrange and Solve for x: Rearrange the equation to set it to zero and solve for x. This means we will subtract 10x and add 80 to both sides of the equation.The new equation is 0=x2−12x+10x+41−80, which simplifies to 0=x2−2x−39.
Factor Quadratic Equation: Factor the quadratic equationx2−2x−39. We are looking for two numbers that multiply to −39 and add up to −2. The numbers that satisfy these conditions are −9 and +7, so we can factor the equation as (x−9)(x+7)=0.
Solve for x: Solve for x by setting each factor equal to zero. This gives us two possible solutions for x: x−9=0 or x+7=0. Solving these, we get x=9 or x=−7.
Substitute x for y: Substitute x back into the original equation y=10x−80 to find the corresponding y values.First, for x=9, we get y=10(9)−80, which simplifies to y=90−80, so y=10.
Find y Values: Now, for x=−7, we substitute into y=10x−80 to get y=10(−7)−80, which simplifies to y=−70−80, so y=−150.
Final Solutions: We now have two solutions for the system of equations: (9,10) and (−7,−150). These are the coordinates of the points where the two equations intersect.
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